Page:The Construction of the Wonderful Canon of Logarithms.djvu/115

Notes. 91 10,000,000 he multiplied his numbers and logarithms by that amount, thereby making them integral to as many places as he intended to print. In this we follow his example, omitting, however, from the formula the indication of this multiplication.

In sec. 30, Napier shows that the logarithm of 9999999, the first proportional after radius in the First table, lies between the limits 1.000000100000010 etc., and 1.000000000000000 etc. And in sec. 31, he proposes to take 1.00000005, the arithmetical mean between these limits, as a sufficiently close approximation to the true logarithm; for, the difference of this mean from either limit being .00000005, it cannot differ from the true logarithm by more than that amount, which is the twenty millionth part of the logarithm, But there can be little doubt that Napier was able to satisfy himself that the difference would be very much less, and that his published logarithms would be unaffected.

We proceed to show the precise amount of error thus introduced into the logarithm of 9999999. If we employ the formula

substituting 10000000 for n, and multiplying the result by 10000000, as before explained, we have

Again, if we take the arithmetical mean of the limits, carried to a similar number of places, we have

The error introduced is consequently

or about a six hundred billionth part in excess of the true logarithm. It will be observed that besides being very much less, this error is in the opposite direction from that caused by the mistake in the Second table.

We have given above the analytical expression for the true logarithm, namely, $$\scriptstyle\frac{1}{n}+\frac{1}{2n^2}+\frac{1}{3n^3}+\frac{1}{4n^4}+\frac{1}{5n^5}+etc.$$ The corresponding expression for the arithmetical mean is $$\scriptstyle\frac{1}{n}+\frac{1}{2n^2}+\frac{1}{2n^3}+\frac{1}{2n^4}+\frac{1}{2n^5}+etc.$$. The latter, therefore, exceeds the true logarithm by $$\scriptstyle\frac{1}{6n^3}+\frac{1}{8n^4}+\frac{1}{10n^5}+etc.$$, which multiplied by n gives $$\scriptstyle\frac{1}{6n^2}+etc.$$ or $$\scriptstyle\frac{1}{6\left(10000000\right)^2}+etc.$$, for the error in Napier’s logarithm. So that up to the 15th place the logarithm obtained